let $f(x)$ is real polynomial. Can we say that $f(\left| x \right|)$ is subharmonic?

Question

A continuous function $\varphi :\mathbb{R} \to \mathbb{C}$ is subharmonic if and only if, for any closed disc in $U$ with centre $\lambda_0$ and radius $r$,$$\varphi ({\lambda _0}) \le \frac{1}{{2\pi }}\int_0^{2\pi } {\varphi ({\lambda _0} + r{e^{i\theta }})d\theta } $$

Now let $f(x)$ is real polynomial.

Can we say that $f(\left| x \right|)$ is subharmonic?

Answer 1

Let $f(x)=1-x$, so $\varphi(x)=1-|x|$, $\lambda=0$, $r=1$. Then $\varphi(0)=1$, but $$ \int_0^{2\pi}\varphi(\lambda_0+e^{i\theta})\,\mathrm d\theta=0$$